Question

don't post irrelevant answers

Answer 1

Answer:

ln3/2

Step-by-step explanation:

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Answer 2

✪ᴀɴsᴡᴇʀ✪

$\boxed{\red{ \lim_{n\to \infty} \sum\limits_{r=0}^{2n} \frac{r}{{n} ^{2} + {r} ^{2}} =\frac{1}{2}ln(5)}}$

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

$\lim_{n\to \infty} \sum\limits_{r=0}^{2n} \frac{r}{{n} ^{2} + {r} ^{2}}$

$=\lim_{n\to \infty} \sum\limits_{r=0}^{2n} \frac{r}{{n} ^{2}(1 + {(\frac{r}{n}) } ^{2})}$

$=\lim_{n\to \infty} \frac{1}{n} \sum\limits_{r=0}^{2n} \frac{\frac{r}{n}}{(1 + {(\frac{r}{n}) } ^{2})}$

$=\int_{0}^{2} \frac{xdx}{(1 + {x}^{2})}$

$=\frac{1}{2} \int_{0}^{2} \frac{2xdx}{(1 + {x}^{2})}$

$=\frac{1}{2}[ln(1+{x}^{2}) ]_{0}^{2}$

$=\frac{1}{2}[ln(1+{2}^{2}) -ln(1+{0}^{2})]$

$=\frac{1}{2}[ln(5) -ln(1)]$

$=\frac{1}{2}[ln(5) -0]$

$=\frac{1}{2}ln(5)$